Topic 6.3: Trigonometry — Exact Trigonometric Values

01 The \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle

An isosceles right-angled triangle with equal legs \(1\) has hypotenuse \(\sqrt{2}\) by Pythagoras. Every exact value at \(45^\circ\) comes from this one diagram.

\[ \sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \qquad \cos 45^\circ = \frac{\sqrt{2}}{2} \qquad \tan 45^\circ = 1 \]

02 The \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle

Cut an equilateral triangle of side \(2\) in half. The altitude is \(\sqrt{3}\) and half the base is \(1\). That is the \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle with sides \(1\), \(\sqrt{3}\), \(2\).

\[ \sin 30^\circ = \tfrac{1}{2} \qquad \cos 30^\circ = \tfrac{\sqrt{3}}{2} \qquad \tan 30^\circ = \tfrac{1}{\sqrt{3}} = \tfrac{\sqrt{3}}{3} \]

\[ \sin 60^\circ = \tfrac{\sqrt{3}}{2} \qquad \cos 60^\circ = \tfrac{1}{2} \qquad \tan 60^\circ = \sqrt{3} \]

Sine and cosine swap when you swap \(30^\circ\) and \(60^\circ\); tangent becomes its reciprocal.

03 The table to learn

You must know \(\sin x\) and \(\cos x\) at \(0^\circ\), \(30^\circ\), \(45^\circ\), \(60^\circ\) and \(90^\circ\), and \(\tan x\) at \(0^\circ\), \(30^\circ\), \(45^\circ\) and \(60^\circ\). \(\tan 90^\circ\) is undefined — it is not on the list.

04 Unit circle (first quadrant)

On a circle of radius \(1\), the point at angle \(\theta\) from the positive \(x\)-axis is \((\cos\theta,\;\sin\theta)\). That is why \(\cos 0^\circ = 1\) and \(\sin 90^\circ = 1\).